Physics
posted by RW .
A jogger travels a route that has two parts. The first is a displacement of 2.90 km due south, and the second involves a displacement that points due east. The resultant displacement + has a magnitude of 4.25 km. (a) What is the magnitude of , and what is the direction of + as a positive angle relative to due south? (b) Suppose that  had a magnitude of 4.25 km. What then would be the magnitude of , and what is the direction of  relative to due south?

X = ?
Y = 2.9 km.
R = 4.25 km = Resultant.
X^2 + Y^2 = R^2
X^2 + (2.9)^2 = (4.25)^2
X^2 = (4.25)^2  (2.9)^2 = 9.653
X = 3.1 km
tanA = 2.9/3.1 = 0.93548
A = 43.1 o
A = 43.1 + 360 = 316.9o CCW.
A = 316.9  270 = 46.9o East of South.